If two torsion theories on a ring localize the ring to the same extension ring, I can find no reason that their "meet" in the lattice of torsion theories must also localize to the same ring. I cannot find anything in Golan's encyclopedia that addresses questions like this.

Does anyone have a counter-example?

Here is a weaker question, not directly related to torsion theories.

Is there an example of the following:

A ring homomorphism R $\to$ S , S-modules P and Q , R-monomorphisms M $\to$ P and M $\to$ Q such that the image of each is an essential R-submodule, but such that the image of M in P $\times$ Q has no essential extension within the product that is an S-submodule

When you say "the same" extension ring, do you mean that the localizations are isomorphic (perhaps in a way that preserves the homomorphism from the original ring into the localization)?
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Manny ReyesDec 4 '12 at 16:15

I had in mind simply isomorphic. But I have no reason to think the property of the meet holds even if the isomorphism is compatible with the structural localization homomorphisms
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Carl WeismanDec 4 '12 at 16:47

Can you remind me what the lattice structure on the torsion theories is? Is it the same as the inclusion ordering on the associated Gabriel filters? I don't have Golan's book handy.
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Manny ReyesDec 7 '12 at 15:09

I believe the lattice structure is indeed the inclusion ordering on Gabriel filters, and the greatest lower bound corresponds to the intersection of the filters.
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Carl WeismanDec 8 '12 at 13:46

The localisations are given by $pr_1 : R \rightarrow k = R_{\mathfrak{F}_1} = eR$ and $pr_2 : R \rightarrow k = R_{\mathfrak{F}_{2}} = (1-e)R$,
but the localisation for $\mathfrak{F}_0$ is given by $id: R = R_{\mathfrak{F}_0}$.

Remark that the $R_{\mathfrak{F}_i}$ for $i =1,2$ really are just isomorphic as rings. There is no isomorphism compatible with the localisations, in fact they are not isomorphic as $R$-modules.

If, on the other hand, we have an isomorphism $j: R_{\mathfrak{F}_1} \xrightarrow{\sim} R_{\mathfrak{F}_2}$ such that for the localisations $\psi_i: R \rightarrow R_{\mathfrak{F}_i}$ we have $j \circ \psi_1 = \psi_2$, then I think the meet will be isomorphic, i.e. we then have $R_{\mathfrak{F}_1 \wedge \mathfrak{F}_2} \simeq R_{\mathfrak{F}_i}$. Here is a sketch of a proof.

First note that because $t_i(R) = ker (\psi_i)$, we have $t_1(R) = t_2(R) =: t(R)$. Set $\bar R := R/t(R)$ and fix an injective hull $E(\bar R)$ of the right $R$-module $\bar R$. It is known that

As you mention Golan, I guess that all your torsion theories are hereditary. Let $\tau_1$ and $\tau_2$ be t.t. on Mod$(R)$ and $\phi_1:R\to R_1$, $\phi_2:R\to R_2$ the two loc. of $R$. The fact that there exists an isomorphism $\phi:R_1\to R_2$ s.t. $\phi\phi_1=\phi_2$, means that $-\otimes_RR_1$ is naturally eq. to $-\otimes_RR_2$. If $\tau_1$ and $\tau_2$ are perfect then these functors coincide with the localization functors. Thus, in such case, $M\in \mathcal T_{\tau_1}$ (the torsion class of $\tau_1$) iff $M\otimes_RR_1=0$ iff $M\otimes_RR_2=0$ iff $M\in \mathcal T_{\tau_2}$. So $ \mathcal T_{\tau_1}=\mathcal T_{\tau_2}$, that is, $\tau_1=\tau_2$.

If your torsion theories are not perfect I do not remember if $\ker(-\otimes_RR_1)=\mathcal T_1$ holds true, if so you should be able to proceed as above...

An equivalent definition of a perfect torsion theory is that localization is equivalent to base change.
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Carl WeismanDec 15 '12 at 13:52

yes, I was not saying that the localization is equivalent to $-\otimes_RR_1$, I was just saying that maybe, when these two functors are different, maybe they have the same kernel... do you have some easy counterexample?
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Simone ViriliDec 15 '12 at 14:16

Let $R& = $K[X,Y]$ for a field K. Let $F$ be the Gabriel filter whose only member is the unit ideal. Let $G$ be the one comprising all ideals not contained in a proper principal ideal. Both localize $R$ to itself. But, while $F$ localizes everything to itself, $G$ localizes $K$, as an $R$-module, to $0$.
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Carl WeismanDec 17 '12 at 13:41

Let R = K[X,Y] for a field K. Let F be the Gabriel filter whose only member is the unit ideal. Let G be the one comprising all ideals not contained in a proper principal ideal. Both localize R to itself. But, while F localizes everything to itself, G localizes K, as an R-module, to 0.
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Carl WeismanDec 17 '12 at 13:45